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For $a\ge \frac{1}{8}$, we define, $$g(a)=\sqrt[\Large3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[\Large3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}$$

Find the maximum value of $g(a)$.

I came across this question in a Math Olympiad Competition and I am not sure how to solve it. Can anyone help? Thanks.

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  • $\begingroup$ I've made an edeit to your post. I believe this is what you ment? If not correct me.... $\endgroup$ – gebruiker Jun 6 '14 at 13:52
  • $\begingroup$ Due to the complexity of the equation(hard to type), I decided to use an image instead. $\endgroup$ – snivysteel Jun 6 '14 at 13:53
  • $\begingroup$ Very smart. I'll try to get is fixed based on the image... $\endgroup$ – gebruiker Jun 6 '14 at 13:53
  • $\begingroup$ I fixed it, the image was a little hard to decipher, is it ok now? $\endgroup$ – Pranav Arora Jun 6 '14 at 13:55
  • $\begingroup$ Yeah, thanks for your help! $\endgroup$ – snivysteel Jun 6 '14 at 13:58
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In general we cannot rewrite a cubic radical in a simple way. However both radicals in $g(a)$ are of a special form that can be denested, because we can find two cubic powers $X^3,Y^3$ such that $X, Y $ are quadratic conjugate irrationals and

\begin{equation*} a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}=X^{3},\qquad a-\frac{a+1}{3}\sqrt{\frac{ 8a-1}{3}}=Y^{3}. \end{equation*}

If we write $x=\frac{8a-1}{3}\geq 0$, then $a=\frac{3x+1}{8}$ and $\frac{a+1 }{3}=\frac{x+3}{8}$. Consequently, the first radicand becomes

\begin{eqnarray*} a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}} &=&\frac{3x+1+\left( x+3\right) \sqrt{x} }{8}=\frac{1+3\sqrt{x}+3x+\left( \sqrt{x}\right) ^{3}}{8} \\ &=&\frac{\left( 1+\sqrt{x}\right) ^{3}}{2^{3}}=X^{3},\qquad X=\frac{ 1+\sqrt{x} }{2}=\frac {1}{2}+\frac {1}{2}\sqrt {\frac {8a-1}{3}}, \end{eqnarray*}

where we used the binomial theorem for the cubic power

\begin{equation*} \left( 1+c\right) ^{3}=1+3c+3c^{2}+c^{3}, \end{equation*}

with $c=\sqrt{x}$:

\begin{eqnarray*} \left( 1+\sqrt{x}\right) ^{3} &=&\left( 1+x^{1/2}\right) ^{3}=1+3x^{1/2}+3x+x^{3/2} \\ &=&1+3\sqrt{x}+3x+\left( \sqrt{x}\right) ^{3}. \end{eqnarray*}

Similarly, using the binomial theorem for $(1-c)^3=(1-\sqrt{x})^3$, the second radical becomes

\begin{eqnarray*} a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}} &=&\frac{3x+1-\left( x+3\right) \sqrt{x} }{8}=\frac{ 1-3\sqrt{x}+3x-\left( \sqrt{x}\right) ^{3}}{2^{3}} \\ &=&\frac{\left( 1-\sqrt{x}\right) ^{3}}{2^{3}}=Y^{3},\qquad Y=\frac{ 1-\sqrt{x} }{2}=\frac {1}{2}-\frac {1}{2}\sqrt {\frac {8a-1}{3}}. \end{eqnarray*}

Now we find easily that for $a\geq 1/8$ the function $g(a)$ is constant

\begin{equation*} g(a)=\sqrt[3]{X^{3}}+\sqrt[3]{Y^{3}}=X+Y=\frac{ 1+\sqrt{x} }{2} +\frac{ 1-\sqrt{x} }{2} =1. \end{equation*}

Hence $\max_{a\geq 1/8}g(a)=1.$

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Hint: Let $u =\sqrt{\frac{8a-1}{3}}$ which implies $a = \frac{3u^2+1}{8}$. Then you have to find the max of \begin{align*} \sqrt[3]{\frac{3u^2+1+u^3+3u}{8}}+\sqrt[3]{\frac{3u^2+1-u^3-3u}{8}}. \end{align*} Observe the expansions of $(1+u)^3$ and $(1-u)^3$.

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Note the $$(a+b)^3=a^3+b^3+3ab(a+b)$$ so $$g^3(a)=2a+3\sqrt[3]{a^2-\dfrac{(a+1)^2}{9}\cdot\dfrac{8a-1}{3}}\cdot g(a)$$

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  • 2
    $\begingroup$ How would you continue? $\endgroup$ – Calvin Lin Jun 6 '14 at 14:58

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